Can you calculate the weight of a sphere of 9 m of displaced water ?
No ? Well, it is 382 tons.
So, the concrete sphere is already massive by itself. "You" don't need any complicated anchoring.
Same goes with the rest of your mechanical engineering intuitions : you did not work in this domain or study it, did you 😆 ?
Also, stress cycling is bad on most material, yes. But here it is compressive stress and the geometry is symmetric. Without further study, i want to believe this thing has good potential and my intuitions tells me it looks nice. Time will tell 😁 !

[–] towerful@programming.dev 12 points 1 day ago (1 children)

Can you calculate the weight of a sphere of 9 m of displaced water ?
No ? Well, it is 382 tons.

Metric strikes again.
I bet you didn't even have to convert through football fields, elephants, or olympic sized swimming pools!

[–] A_A@lemmy.world 3 points 1 day ago

indeed i made a very simplified calculation not taking into account increase density of salted water nor increased density because of compressibility of water at 500 m deep. Basically i took 1m³(water) is 1 (metric) ton.

[–] carl_dungeon@lemmy.world 9 points 1 day ago (1 children)

Thanks for the insight, I’m not a mechanical engineer, I’m a software engineer :) The walls on these spheres have got to be pretty thick- 400 tonnes is no joke. 3/4 of a meter if I had to guess.

[–] A_A@lemmy.world 9 points 1 day ago (1 children)

Perfect guess ! (afaik) ρ(concrete) ≈ 2.5 tons/m³
so full sphere ≈ (2.5 x 382) tons = 955 tons
they have 400 t so the cavity removes :
955 - 400 = 555 t ... so 7.51m diam. cavity
... so, yes 3/4m thick wall 😌👍 !

[–] Cocodapuf@lemmy.world 4 points 8 hours ago* (last edited 8 hours ago)

That's exactly the way I would have calculated it, glad someone beat me to it though. Thanks!